Maths ERGENT HW !!!!!!!!!!!!!!!!20pts

Let f(x) = x^99 +k .
a) When f(x) is divided by x+1, the remainder is 1. Find the values of k

By doing a long division, we can show that x+1 is a factor of x^99+1, namely
x^99+1
=(x+1)*(x^2-x+1)*(x^6-x^3+1)*(x^10-x^9+x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1)*
(x^20+x^19-x^17-x^16+x^14+x^13-x^11-x^10-x^9+x^7+x^6-x^4-x^3+x+1)*
(x^60+x^57-x^51-x^48+x^42+x^39-x^33-x^30-x^27+x^21+x^18-x^12-x^9+x^3+1)

Hence, x^99+2 will give a remainder of 1 when divided by (x+1), or in other words, k=2.

b) Hence, find the remainder when 9^99 is divided by 10

By putting x=9, then 9^99+1 divides exactly by (x+1)=(9+1)=10.
Therefore 9^99 / 10 has a remainder of 10-1=9.

(note: 9^99=
29512665430652752148753480226197736314359272517043832886063884637676943433478020332709411004889)

2007-10-28 11:44:17 補充：
Note for part a)In fact, (x 1) is a factor of x^n 1 where n is odd.So the long division is not really necessary.

a)
By remainder theorem, f(-1) = 1.

f(x) = x^99 +k
f(-1) = (-1)^99 + k
1 = -1 + k
k = 2

b)
By (a), take x = 9,
f(9) = 9^99 + 2
When f(9) is divided by (9 + 1), remainder is 1.
So, when 9^99 is divided by 10, remainder is (1 - 2 = -1) i.e. 9.